Force on a proton in electric field #vecE# is given by
#vecF=qvecE#
Acceleration produced is #veca=q/mvecE#
where #q and m# are the proton's charge and mass.
Treating orthogonal velocities independently.
Time taken by proton to travel horizontal distance #d# to just exit out of electric field
#t=d/v# ......(1)
Vertical component of velocity #v_(xt)# attained in this time is found by the following kinematic expression
#v=u+at#
#v_(xt)=q/m|vecE|d/v#
#v_(xt)=(qd)/(mv)|vecE|# .....(2)
We know that angle of the resultant of two vectors #vecP and vecQ# having angle #theta# between them is given by the formula
#tan alpha=(|vecQ|sintheta)/(|vecP|+|vecQ|costheta)#
We get at the exit point
#tan 15^@=(v_(xt)sin90^@)/(v+v_(xt)cos90^@)#
#=>tan 15^@=(v_(xt))/(v)#
Using (2) we get
#tan 15^@=((qd)/(mv)|vecE|)/(v)#
#=>tan 15^@=(qd|vecE|)/(mv^2)#
#=>|vecE|=(mv^2tan 15^@)/(qd)# .....(3)
Taking charge of proton #q=1.602xx10^-19 C#,
mass of proton #m=1.673 xx 10^-27 kg#
#|vecE|=932.7Vm^-1#
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
Alternate method after equation (1).
During time #t# vertical deflection produced to meet the condition is
#"deflection"/(d/2)=tan15^@#
as the line of exit velocity extended backwards meets the axis of velocity #vecv# at mid-point of #d# i.e., at a distance #=d/2#
Setting up the kinematic equation for vertical deflection and noting that initial vertical velocity of proton is zero as it enters the uniform electric field region
#s=ut+1/2at^2#
#d/2tan 15^@=1/2(q/m|vecE|)(d/v)^2#
#d tan 15^@=(q/m|vecE|)(d/v)^2#
#=>|vecE|=m/qd tan15^@(v/d)^2#
#=>|vecE|=(mv^2tan15^@)/(qd)#
which is same as equation (3)
